Deepthi Subbu wrote:One of the components of an industrial machine is a wire loop pictured here. It consists of a metal circle and a straight 'tail.' These loops fit perfectly in cylindrical containers with a capacity of 32Ï€ in3. How much wire must be used to create each loop?
(1) The height of each container is 8 in.
(2) The diameter of the loop is 4 in.
As the loops fit perfectly inside the container, height of the container is same as length of the tail of the loop and radius of the container is same as radius of the loop.
Say, radius of the cylindrical container is r and height is h.
Then its volume = πr²h = 32π
=> r²h = 32
Then, radius of the loop is r and length of the tail is h.
Then, the length of wire used to make the loop = (2Ï€r + h)
Therefore we need to determine h and r.
Statement 1: Height of each container is 8 in.
Hence, h = 8
=> r² = (32/8) = 4
=> r = 2
Sufficient.
Statement 2: The diameter of the loop is 4 in.
Hence, r = 2
=> r² = 4
=> h = (32/4) = 8
Sufficient
The correct answer is D.
